Question

A projectile is shot at an angle of `pi/12` and a velocity of `3 2 m/s`. How far away will the projectile land?

Answer 1

The distance is `=52.25m`

Explanation:

The equation describing the trajectory of the projectile in the `x-y` plane is

`y=xtantheta-(gx^2)/(2u^2cos^2theta)`

The initial velocity is `u=32ms^-1`

The angle is `theta=(1/12pi)rad`

The acceleration due to gravity is `g=9.8ms^-2`

The distance `y=0`

Therefore,

`xtan(1/12pi)-(9.8*x^2)/(2*32^2cos^2(1/12pi))=0`

`0.268x-0.0051x^2=0`

`x(0.268-0.0051x)=0`

`x=0`, this is the starting point

`x=0.268/0.0051=52.25m`

graph{0.268x-0.0051x^2 [-5, 68.1, -15.73, 20.8]}